Let $G(A,B)$ be a bipartite graph with $|A| = |B| = n$, where $n$ is sufficiently large. The average degree of $G$ is $d = \frac{e(A,B)}{n}$, where $e(A,B)$ denotes the number of edges between sets $A$ and $B$.
Definition of regularity: Let $0 < \varepsilon \ll d < 1$. We say that $G$ is $(\varepsilon, d)$-regular if for any subset $A' \subseteq A$ and $B' \subseteq B$ with $|A'|, |B'| \geq \varepsilon n$, we have
$$ (d - \varepsilon) |A'| |B'| \leq e(A',B') \leq (d + \varepsilon) |A'| |B'|. $$
Question: If for any subset $A' \subseteq A$ with $|A'| = n/2$ and $e(A',B) = \frac{dn^2}{2}$, and for any subset $B' \subseteq B$ with $|B'| = n/2$ and $e(A,B') = \frac{dn^2}{2}$, we have
$$ (d - \varepsilon) |A'| |B'| \leq e(A',B') \leq (d + \varepsilon) |A'| |B'|, $$
can we conclude that $G$ exhibits some form of regularity (or pseudo-randomness), i.e. $G$ contains a regular subgraph which contains linar numbers of edges with respect to $e(G)$?
In other words, does there exist a constant $c > 0$ (which is independent of $n$, $d$, and $\varepsilon$; for example, $c = 1/4$) and subsetgraph $G'$ with $e(G') \geq c (dn^2)=ce(G)$ such that $G(G')$ is $(f(\varepsilon), d)$-regular, where $f(\varepsilon)$ is some function of $\varepsilon$, possibly $f(\varepsilon) = 4\varepsilon$?
Note that when $G$ is a regular graph (i.e., every vertex has the same degree), the answer is Yes with $c = 1$ and $f(\varepsilon) = \varepsilon$, due to another equivalent definition of regularity (where $A'$, $B'$ can be any subset with $|A'| = |B'| = n/2$). However, when $G$ is not regular, the answer is unclear for me.
